Question

How do you find the integral of \[\int {\dfrac{x}{{{{\cos }^2}x}}dx} \]?

Answer 1

Hint: In order to determine the answer of above definite integral use the formula of integration by parts i.e. $\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} } $ and assume $f(x) = x$ and $g'(x) = {\sec ^2}x$ and calculate $f'(x)$ and $g(x)$ and put into the formula and use the substitution method to find the integral of the second term to get the final result. Don’t forget to place Constant of integration $C$ at the end of the integral.

Formula used:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} $
$\int {\dfrac{1}{x} = \ln x + C} $
\[{\cos ^2}x + {\sin ^2}x = 1\]
$\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} } $

Complete step by step answer:
We are given an expression \[\int {\dfrac{x}{{{{\cos }^2}x}}dx} \] and we have to calculate its integral.
Rewriting the integral using the fact that$\dfrac{1}{{{{\cos }^2}x}} = {\sec ^2}x$
\[I = \int {x{{\sec }^2}x\,dx} \]
To calculate the integral of \[\int {x{{\sec }^2}x\,dx} \],we will be using Integration by parts method
The formula for calculation of integration of parts is
$\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} } $
In our question let’s assume $f(x) = x$ and $g'(x) = {\sec ^2}x$
As we know that the derivative of function $\tan x$ i.e.$\dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$. So $g(x) = \tan x$, and derivative of function x is equal to 1,so \[f'(x) = 1\] now putting the values of $f(x),f'(x),g(x)\,and\,g'(x)$ into the formula of Integration by parts
\[I = \int {x{{\sec }^2}x\,dx} = x\tan x - \int {(1).\tan x\,} dx\]

Substituting $\tan x = \dfrac{{\sin x}}{{\cos x}}$
\[
\Rightarrow I = \int {x{{\sec }^2}x\,dx} = x\tan x - \int {\dfrac{{\sin x}}{{\cos x}}\,} dx \\
 \Rightarrow x\tan x + \int {\dfrac{{ - \sin x}}{{\cos x}}\,} dx \\
 \]
Integral of the 2nd term can be determined using integration by substitution method, Let $y \Rightarrow \sin x$, so $dy = - \sin x dx$.
Substituting $ - \sin xdx = dy\,\,and\,\cos x = y$, we get
\[ \Rightarrow x\tan x + \int {\dfrac{{dy}}{y}\,} \]
Integration of $\dfrac{1}{y} = \ln \left( {\left| y \right|} \right) + C$
\[
\Rightarrow x\tan x + \int {\dfrac{{dy}}{y}\,} \\
 \Rightarrow x\tan x + \ln \left( {\left| y \right|} \right) + C \\
 \]
$\therefore I = \int {x{{\sec }^2}x\,dx} = x\tan x + \ln \left( {\left| y \right|} \right) + C$
C is the constant of integration.

Therefore, the integral \[\int {\dfrac{x}{{{{\cos }^2}x}}dx} \] is equal to $x\tan x + \ln \left( {\left| y \right|} \right) + C$.

Additional Information:
1.Different types of methods of Integration:
Integration by Substitution
Integration by parts
Integration of rational algebraic function by using partial fraction

2. Integration by Substitution: The method of evaluating the integral by reducing it to standard form by a proper substitution is called integration by substitution.
If $\varphi (x)$ is a continuously differentiable function, then to evaluate integrals of the form.
\[\int {f(\varphi (x)){\varphi^1}(x) dx} \], we substitute $\varphi (x)$ = t and ${\varphi ^1}(x)dx = dt$
This substitution reduces the above integral to \[\int {f(t)dt} \]. After evaluating this integral we substitute back the value of t.

Note: 1. Use standard formula carefully while evaluating the integrals.
2. Indefinite integral=Let $f(x)$ be a function .Then the family of all its primitives (or antiderivatives) is called the indefinite integral of $f(x)$ and is denoted by $\int {f(x)} dx$
3. The symbol $\int {f(x)dx} $ is read as the indefinite integral of $f(x)$with respect to x.
4. C is known as the constant of integration.